Đáp án:
1,C
2,C
3,B
Giải thích các bước giải:
1,
\(\begin{array}{l}
HCOO{C_2}{H_5} + NaOH \to HCOONa + {C_2}{H_5}OH\\
{n_{HCOO{C_2}{H_5}}} = 0,05mol\\
\to {n_{HCOONa}} = {n_{HCOO{C_2}{H_5}}} = 0,05mol\\
\to m = {m_{HCOONa}} = 3,4g
\end{array}\)
2,
\(\begin{array}{l}
{n_{C{O_2}}} = 0,26mol\\
{n_{{H_2}O}} = 0,26mol\\
\to {n_{C{O_2}}} = {n_{{H_2}O}} = 0,26mol
\end{array}\)
Suy ra este no, đơn chức, mạch hở
\(\begin{array}{l}
\to {n_C} = {n_{C{O_2}}} = 0,26mol\\
\to {n_H} = 2{n_{{H_2}O}} = 0,52mol\\
{m_X} = {m_C} + {m_H} + {m_O}\\
\to {m_O} = 7,8 - 0,26 \times 12 - 0,52 \times 1 = 4,16g\\
\to {n_O} = 0,26mol\\
\to {n_C}:{n_H}:{n_O} = 0,26:0,52:0,26 = 1:2:1\\
\to {(C{H_2}O)_n}\\
n = 2 \to {C_2}{H_4}{O_2}
\end{array}\)
3,
\(\begin{array}{l}
{n_{C{O_2}}} = 0,15mol\\
{n_{{H_2}O}} = 0,15mol\\
\to {n_{C{O_2}}} = {n_{{H_2}O}} = 0,15mol
\end{array}\)
Suy ra este no, đơn chức, mạch hở
\(\begin{array}{l}
\to {n_C} = {n_{C{O_2}}} = 0,15mol\\
\to {n_H} = 2{n_{{H_2}O}} = 0,3mol\\
{m_X} = {m_C} + {m_H} + {m_O}\\
\to {m_O} = 3,7 - 0,15 \times 12 - 0,3 \times 1 = 1,6g\\
\to {n_O} = 0,1mol\\
\to {n_C}:{n_H}:{n_O} = 0,15:0,3:0,1 = 3:6:2\\
\to {C_3}{H_6}{O_2}
\end{array}\)
